# physics

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A 40 g marble moving at 2.3 m/s strikes a 29 g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on.

What is the speed of the first marble immediately after the collision?

What is the speed of the second marble immediately after the collision?

I've tried multiple ways of setting this up and each time its not right

• physics - ,

The center of mass (CM) moves at a velocity
Vcm = 0.40*2.3/0.69 = 1.333 m/s

In a coordinate system moving with the center of mass, the 40g marble moves at speed
2.3 - 1.333 = 0.967 m/s before collision and -0.967 m/s after collision. In that same coordinate system, the 29 g marble moves at -1.333 m/s before collision and +1.333 m/s after collision.

Now transfer final velocities in CM coordinates back to laboratory coordinates by adding 1.333 m/s to each.
The 40g marble moves at 0.366 m/s and the 29 g marble moves at 2.666 m/s after collision. (The latter is the answer you wanted)

Momentum check (in lab coordinates):
Initial momentum = 0.40*2.3 = 0.920 kg m/s
Final momentum = 0.40*0.366 + 0.29*2.666 = 0.920 m/s

Kinetic energy is also conserved.